# Work done on dipole and potential energy in uniform electric field

• ShaunPereira

#### ShaunPereira

Homework Statement
Finding out relevant equation for work done on a dipole and its potential energy.
Relevant Equations
$$W= -\Delta U$$
$$U= -PEsin\theta$$
I encountered a problem regarding the appropriate sign needed to be taken for the work done on a dipole when it rotates in a uniform electric field and would appreciate some help.

The torque on a dipole can be defined as
τ=PEsinθ
The work done on a dipole to move it from an angle ##\theta_0## to an angle ##\theta_1## can be written as the integral of ##\tau.dr## from ##\theta_0## to ##\theta_1##
which is PE[cos##\theta_0##- ##\theta_1## ]

The change in potential energy would be ##U_2 -U_1##
=##-PEcos\theta_1## -(-##PEcos\theta_0##)
=##PEcos\theta_0##- ##PEcos\theta_1##
=PE[cos##\theta_0##- ##\theta_1## ]

If we compare the equations obtained for work done and the change in potential they both are the same which make sense since the change in potential energy would be the work done.

But this is where my problem begins. Isn't the work done negative of the change in potential energy due to a conservative force?
If yes why is ##W= \Delta U## here and not ##W= -\Delta U##
I know this sound silly but does it make a difference?
I have encountered problems with multiple choices where the sign mattered and wasn't sure which one to choose.

Let's talk about a simpler situation, doing work lifting a brick. If ##F## is the force you apply to the brick, then the work you do on the brick is,

##W = F\cdot \Delta S##

where ##\Delta S## is the vector displacement of the brick. Well, to move the brick up you must push up. Both ##F## and ##\Delta S## are in the same direction so the work is positive. If the brick started at rest and is at rest when you finish lifting it, then ##W## is the change in the potential energy.

If you stop pushing on the brick, it will fall acted upon by a force,

##F = -\nabla U.##

The minus sign makes sense because the force must undo the increase of potential energy put in by lifting.

Let's talk about a simpler situation, doing work lifting a brick. If ##F## is the force you apply to the brick, then the work you do on the brick is,

##W = F\cdot \Delta S##

where ##\Delta S## is the vector displacement of the brick. Well, to move the brick up you must push up. Both ##F## and ##\Delta S## are in the same direction so the work is positive. If the brick started at rest and is at rest when you finish lifting it, then ##W## is the change in the potential energy.

If you stop pushing on the brick, it will fall acted upon by a force,

##F = -\nabla U.##

The minus sign makes sense because the force must undo the increase of potential energy put in by lifting.
Right so if I understood you correctly then what you essentially mean is:
The sign depends on who is doing the work
IF it is by an external agent who lifts the brick, both the the force and the displacement are in the same direction and hence the work done by it is positive and the work done by it is simply the change in potential energy.

However the gravitational force of the Earth on the brick and displacement are in the opposite direction leading to negative work being done by the force which is the negative of change in potential energy.

Please clarify if the above conclusion is right or wrong.

ALSO
and it asked me to find the work done by the external agent to turn the dipole which would essentially be the change in potential energy.

given by $$W=\Delta U$$

Had they asked the work done by the electric field to turn the dipole through some angle I'd assume it would be

$$W=-\Delta U$$

Is this correct?

I think you have it.

• ShaunPereira
Ok. Thank You.